M.C.Q (1 Marks) - Maths STD 10 Questions - Vidyadip

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M.C.Q (1 Marks)

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MCQ 11 Mark

The angle of depression of a car parked on the road from the top of a $150 m$ high tower is $30^{\circ}$. The distance of the car from the tower $($in metres$)$ is

A
$50 \sqrt{3}$
✓
$50 \sqrt{3}$
C
$150 \sqrt{2}$
D
$75$

Answer

Correct option: B.

$50 \sqrt{3}$

Consider the tower $\text{AB}$ of height $150 m$ and $\angle \text{DAC}$
$=\angle A C B=30^{\circ}($ alternate angles $)$
$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$\Rightarrow \tan 30^{\circ}=\frac{AB}{BC}$
Using $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{150}{BC}$
$\Rightarrow BC=150 \sqrt{3} m$
Thus the correct answer is $(b).$

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MCQ 21 Mark

A ladder makes an angle of $60^{\circ}$ with the ground when placed against a w all. If the foot of the ladder is $2\ m$ away from the wall, then the length of the ladder $($in meters$)$ is:

A
$\frac{4}{\sqrt{3}}$
B
$4 \sqrt{3}$
C
$2 \sqrt{2}$
✓
$4$

Answer

Correct option: D.

$4$

Let the length of the ladder is $M N$, placed against the wall $A B$ and makes an angle of $60^{\circ}$ with the ground.
The foot of the ladder is at $N$, which is $2\ m$ away from the wall.
$B N=2\ m$
In right $-$ angled triangle $\triangle M B N$ :
$\cos 60^{\circ}=\frac{B N}{M N}=\frac{2}{M N}$
Using $\cos 60^{\circ}=\frac{1}{2}$ we get,
$\frac{1}{2}=\frac{2}{M N}$
$\Rightarrow M N=4\ m$
Therefore, the length of the ladder is $4\ m$ .

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MCQ 31 Mark

The angle of depression of a car, standing on the ground, from the top of a $75 m$ high tower, is $30^{\circ}$. The distance of the car from the base of the tower $($in $m .)$ is:

A
$25 \sqrt{3}$
B
$50 \sqrt{3}$
✓
$75 \sqrt{3}$
D
150

Answer

Correct option: C.

$75 \sqrt{3}$

Let $A B$ is the Tower of height $75 m$ and Car is at the point $C$ on the ground.
Now,
According to the question,
$\angle \text{CBD}=30^{\circ}$
$\angle \text{BCA}=30^{\circ} ($alternate opposite angles $)$
And,
In the $\triangle \text{ABC}$,
$\operatorname{Cot} 30^{\circ}=\frac{A C}{A B}$
$\Rightarrow A C=A B \operatorname{Cot} 30^{\circ}$
Putting $\operatorname{Cot} 30^{\circ}=\sqrt{3}$
$A C=75 \times \sqrt{3}$
$\therefore$ The distance of the car from the base of the tower is $75 \sqrt{3} m$
Hence option $(c)$ is correct.

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MCQ 41 Mark

A kite is flying at a height of $30\ m$ from the ground. The length of string from the kite to the ground is $60
\ m$ . Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is

A
$45^{\circ}$
✓
$30^{\circ}$
C
$60^{\circ}$
D
$90^{\circ}$

Answer

Correct option: B.

$30^{\circ}$

$\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}$
$\Rightarrow \sin \theta=\frac{30}{60}$
$\Rightarrow \sin \theta=\frac{1}{2}$
$\Rightarrow \sin \theta=\sin 30^{\circ}$
$\Rightarrow \theta=30^{\circ}$
The angle of elevation of kite at the ground is $30^{\circ}$.

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MCQ 51 Mark

The angle of elevation of the top of a tower from a point on the ground, which is $30\ m$ away from the foot of the tower is $45^{\circ}$. The height of the tower $($in metres$)$ is

A
$15$
✓
$30$
C
$30 \sqrt{3}$
D
$10 \sqrt{3}$

Answer

Correct option: B.

$30$

In the figure above, $AB$ is the tower and $C$ is the point $30 \ m$ away from the foot of the tower.
Let $h$ denote the height of the tower $($in metres$).$
$\tan 45^{\circ}=\frac{AB}{BC}=\frac{h}{30}$
$\Rightarrow 1=\frac{h}{30}$
$\Rightarrow h=30$

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MCQ 61 Mark

In figure, $\text{PA , QB}$ and $\text{RC}$ are each perpendicular to $\text{AC}$ . If $x=8 \ cm$ and $z=6 \ cm$, then $y$ is equal to

A
$\frac{56}{7} cm$
B
$\frac{7}{56} cm$
C
$\frac{25}{7} cm$
✓
$\frac{24}{7} cm$

Answer

Correct option: D.

$\frac{24}{7} cm$

$\Rightarrow \frac{BQ}{AP}=\frac{CB}{CA}$
[Since, $\triangle \text{CBQ} \sim \Delta \text{CA} ]$
$\Rightarrow \frac{y}{x}=\frac{BQ}{CR} \ldots \ldots(i)$
In $\triangle \text{ACR}$, we have $B Q|\mid C R$
$\Rightarrow \frac{BQ}{AP}=\frac{AB}{AC}$
$[$Since, $\triangle \text{ABQ} \sim \triangle \text{ACR} ]$
$\Rightarrow \frac{y}{x}=\frac{AB}{AC}$
Adding $(i)$ and $(ii)$, we get
$\frac{y}{x}+\frac{y}{z}=\frac{CB}{AC}+\frac{AB}{AC}$
$\Rightarrow \frac{y}{x}+\frac{y}{z}=\frac{AB+BC}{AC}$
$\Rightarrow \frac{y}{x}+\frac{y}{z}=\frac{AC}{AC}$
$\Rightarrow \frac{y}{x}+\frac{y}{z}=1$
$\Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{1}{y}$
Put $x=8$ and $z=6$
$\frac{1}{y} =\frac{1}{8}+\frac{1}{6}=\frac{14}{48}=\frac{7}{24}$
$ \Rightarrow y=\frac{24}{7}$

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MCQ 71 Mark

The angle of elevation of the top of a 15 m high tower at a point $15 \sqrt{3} m$ away from the base of the tower is:

✓
$30^{\circ}$
B
$45^{\circ}$
C
$60^{\circ}$
D
$90^{\circ}$

Answer

Correct option: A.

$30^{\circ}$

(a)
$\tan 30^{\circ}=\frac{1}{\sqrt{3}}=30^{\circ}$

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MCQ 81 Mark

A pole casts a shadow of length $2 \sqrt{3} m$ on the ground when the sun's elevation is $60^{\circ}$. The height of the pole is

A
$4 \sqrt{3} m$
✓
$6 m$
C
$12 m$
D
$3 m$

Answer

Correct option: B.

$6 m$

Let the height of the pole be $h$ metres

Then, $\tan 60^{\circ}=\frac{ h }{2 \sqrt{3}}$
$=\sqrt{3}=\frac{h}{2 \sqrt{3}}$
$\Rightarrow h=2 \times 3$
$=6 m$

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MCQ 91 Mark

The length of vertical rod and its shadow are in the ratio $1: \sqrt{3}$. The angle of elevation of the sun is

✓
$30^{\circ}$
B
$45^{\circ}$
C
$60^{\circ}$
D
$90^{\circ}$

Answer

Correct option: A.

$30^{\circ}$

Let $A B$ be the rod of length $x$ metres and let $A C$ be its shadow of length $\sqrt{3} xm$.

Let $\angle ACB =\theta$.
Then, $\frac{ AB }{ AC }=\tan \theta$
$\Rightarrow \tan \theta=\frac{x}{\sqrt{3} x}=\frac{1}{\sqrt{3}}$
$\Rightarrow \tan \theta=\tan 30^{\circ} $
$\Rightarrow \theta=30^{\circ}$

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MCQ 101 Mark

The angle of depression of a car parked on the road from the top of a $150 m$ high tower is $30^{\circ}$. The distance of the car from the tower is.

A
$50 \sqrt{3} m$
✓
$150 \sqrt{3} m$
C
$150 \sqrt{2} m$
D
75 m

Answer

Correct option: B.

$150 \sqrt{3} m$

Let $A B$ be the tower and $B C=30 m$ be the ground such that $\angle \text{BCA} =30^{\circ}$

Let $A C=x m$.
Then, $\frac{AC}{AB}=\cot 30^{\circ}$
$\Rightarrow \frac{x}{150}=\sqrt{3}$
$\Rightarrow x=150 \sqrt{3} m$

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MCQ 111 Mark

From a point on the ground, $30 m$ away from the foot of a tower, the angle of elevation of the top of the tower is $30^{\circ}$. The height of the tower is

A
$30 m$
✓
$10 \sqrt{3} m$
C
$10 m$
D
$30 \sqrt{3} m$

Answer

Correct option: B.

$10 \sqrt{3} m$

Let $\text{AB}$ be the tower and $\text{BC}=30 m$ be the ground such that $\angle \text{BCA} =30^{\circ}$

Let $A B=h m$, Then,
$\frac{AB}{BC}=\tan 30^{\circ}$
$\Rightarrow \frac{h}{30}=\frac{1}{\sqrt{3}}$
$h=\frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$=10 \sqrt{3} m$

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MCQ 121 Mark

A ladder $15 m$ long just reaches the top of a vertical wall. If the ladder makes an angle of $60^{\circ}$ with the wall then height of the wall is

A
$15 \sqrt{3} m$
B
$\frac{15 \sqrt{3}}{2} m$
C
$15 m$
✓
$\frac{15}{2} m$

Answer

Correct option: D.

$\frac{15}{2} m$

Let $AB$ be the ladder and $BC$ be the wall.

Then,
$\angle \text{ABC}=60^{\circ}$
$\Rightarrow \angle \text{CAB}=\left(90^{\circ}-60^{\circ}\right)=30^{\circ}$
Let $\text{BC}=h m$. Then,
$\frac{BC}{AB}=\sin 30^{\circ}$
$\Rightarrow \frac{h}{15}=\frac{1}{2}$
$\Rightarrow h=\frac{15}{2} m$
Hence, the height of the wall $=\frac{15}{2} m$

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MCQ 131 Mark

A ladder makes an angle of $60^{\circ}$ with the ground placed against a wall. If the foot of the ladder is $2m$ away from the wall, the length of the ladder is

A
$\frac{4}{\sqrt{3}} m$
B
$4 \sqrt{3} m$
C
$2 \sqrt{2} m$
✓
$4 m$

Answer

Correct option: D.

$4 m$

Let $\text{AB}$ be the ladder and $\text{BC}$ be the wall.

Then $\angle \text{CAB} =60^{\circ}$ and $\text{CA} =2 m$.
Let $A B=x m$. Then,
$\frac{AC}{AB}=\cos 60^{\circ}$
$\Rightarrow \frac{2}{x}=\frac{1}{2}$
$\Rightarrow x=4$
So, the length of the ladder $=4 m$

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MCQ 141 Mark

The shadow of a $5m$ long stick is $2 m$ long. At the same time, the length of the shadow of a $12.5 m$ high tree is

A
$3 m$
B
$3.5 m$
C
$4.5 m$
✓
$5 m$

Answer

Correct option: D.

$5 m$

Ratio of lengths of objects $=$ ratio of lengths of their shadows.
Let the length of shadow of the tree be xm.
Then, $\frac{5}{12.5}=\frac{2}{x}$
$\Rightarrow 5 x=2 \times 12.5$
$\Rightarrow x=\frac{2 \times 12.5}{5}=5$
So, length of shadow of the tree $=5 m$.

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MCQ 151 Mark

If a pole $12 m$ high casts a shadow $4 \sqrt{3}$ long on the ground then the sun's elevation is

✓
$60^{\circ}$
B
$45^{\circ}$
C
$30^{\circ}$
D
$90^{\circ}$

Answer

Correct option: A.

$60^{\circ}$

Let $A B$ be the pole and $A C$ be its shadow. $AB =12 n$, and $AC =4 \sqrt{3} m$

Let $\angle \text{ACB} =\theta$. Then,
$\frac{AB}{AC}=\frac{12}{4 \sqrt{3}}$
$\Rightarrow \tan \theta=\frac{12}{4 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\sqrt{3}$
$\Rightarrow \tan \theta=\tan 60^{\circ}$
$\Rightarrow \theta=60^{\circ}$

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MCQ 161 Mark

If the height of a vertical pole is $\sqrt{3}$ times the length of its shadow on the ground then the angle of elevation of the sun at this time is

A
$30^{\circ}$
B
$45^{\circ}$
✓
$60^{\circ}$
D
$75^{\circ}$

Answer

Correct option: C.

$60^{\circ}$

Let $A B$ be the pole and $A C$ be its shadow.
Let $AC = xm$.
Then, $AB =\sqrt{3} xm$.
Let $\angle ACB =\theta^{\circ}$.

Then, $\tan \theta=\frac{ AB }{ AC }=\frac{\sqrt{3} x }{ x }$
$\Rightarrow \tan \theta=\sqrt{3}$
$\Rightarrow \tan \theta=\tan 60^{\circ}$
$\Rightarrow \theta=60^{\circ}$

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MCQ 171 Mark

If the height of a vertical pole is equal to the length of its shadow on the ground, the angle of elevation of the sun is

A
$0^{\circ}$
B
$30^{\circ}$
✓
$45^{\circ}$
D
$60^{\circ}$

Answer

Correct option: C.

$45^{\circ}$

Let $A B$ be the pole and $A C$ be its shadow such that $AB = AC$

Let $\angle ACB =\theta$. Then
$\tan \theta=\frac{A B}{A C}=1$
$\tan \theta=\tan 45^{\circ}$
$\therefore \theta=45^{\circ}$

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